Problem: Evaluate the improper integral if it exists. $\int_{0}^{1}\dfrac{1}{\sqrt{x}}\,dx $ Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $2$ (Choice C) C $4$ (Choice D) D The improper integral diverges.
First, let's rewrite the improper integral: $\int_{0}^{1}\dfrac{1}{\sqrt{x}}\,dx =\lim_{a\to0^+}\int_a^{1} \dfrac{1}{\sqrt{x}}\,dx $ We can now evaluate the integral: $\begin{aligned} \phantom{\int_0^{1}\dfrac1{\sqrt{x}}dx}&=\lim_{a\to0^+}\int_a^1 \dfrac1{\sqrt{x}}dx\\ \\ \\ &=\lim_{a\to0^+}\Big[2\sqrt{x}\Big]_a^1\\ \\ \\ &=2\lim_{a\to0^+}\left(\sqrt{1}-\sqrt{a}\right)\\ \\ \\ &=2\left(\lim_{a\to0^+}1-\lim_{a\to0^+}\sqrt{a}\right)\\ \\ &=2(1-0)\\ \\ &=2 \end{aligned}$ The answer: $\int_{0}^{1}\dfrac{1}{\sqrt{x}}\,dx =2$